First slide

Properties of ITF

Question

$If \tan^{- 1} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}} = α,$ then x² =

Moderate

A

cos 2α
B

sin 2α
C

tan 2α
D

cot 2α

Solution

$\begin{array}{l} Since, \tan^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}) = α \\ \Rightarrow \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}} = \frac{\tan α}{1} \end{array}$
Using componendo and dividendo, we get
$\begin{array}{l} \frac{2 \sqrt{1 + x^{2}}}{2 \sqrt{1 - x^{2}}} = \frac{1 + \tan α}{1 - \tan α} = \tan (\frac{π}{4} + α) \\ \Rightarrow \frac{1 + x^{2}}{1 - x^{2}} = \frac{\tan^{2} (\frac{π}{4} + α)}{1} \end{array}$
Again using componendo and dividendo, we have
$\begin{array}{l} \frac{(1 + x^{2}) - (1 - x^{2})}{(1 + x^{2}) + (1 - x^{2})} = \frac{\tan^{2} (π / 4 + α) - 1}{\tan^{2} (π / 4 + α) + 1} \\ \Rightarrow x^{2} = - \cos (\frac{π}{2} + 2 α) = \sin 2 α \end{array}$

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